The Problem Statement: 12V to 5.4V Series Dropper Network
Given: You have a 12.0V DC bench supply powering a high-impedance logic circuit. You need to design an example of a series circuit using three resistors ($R_1$, $R_2$, $R_3$) to drop the voltage to exactly 5.4V across $R_3$ (which feeds the logic input). To minimize standby power draw and thermal noise, the total current drawn from the supply must be exactly 2.0 mA.
Fixed Components: $R_1$ is a 1.8 kΩ precision resistor. $R_2$ is a 1.5 kΩ precision resistor.
Tasks:
- Calculate the exact required resistance for $R_3$.
- Calculate the individual voltage drops across $R_1$ and $R_2$.
- Select the nearest standard E12 series resistor for $R_3$.
- Determine the physical wattage rating required for $R_3$ using a 2x safety margin, and specify the final part to buy.
Method Selection and the Common Wattage Trap
Because this is a pure series circuit, the current is constant through all components. Therefore, Ohm's Law ($V = IR$) and Kirchhoff's Voltage Law (KVL) are the governing theorems. KVL dictates that the sum of the voltage drops across the series resistors must equal the source voltage. As detailed in standard circuit theory references like All About Circuits, the equivalent resistance of a series network is simply the arithmetic sum of its parts.
Step-by-Step Algebraic Solution
We will solve this systematically, showing every algebraic step to ensure no assumptions are skipped.
Step 1: Find the total required circuit resistance ($R_{total}$)
Using Ohm's Law for the entire circuit:
$R_{total} = \frac{V_{source}}{I_{total}}$
$R_{total} = \frac{12.0\text{ V}}{0.002\text{ A}}$
$R_{total} = 6000\text{ }\Omega = 6.0\text{ k}\Omega$
Step 2: Solve for the exact value of $R_3$
Using the series resistance formula:
$R_{total} = R_1 + R_2 + R_3$
$6000\text{ }\Omega = 1800\text{ }\Omega + 1500\text{ }\Omega + R_3$
$6000\text{ }\Omega = 3300\text{ }\Omega + R_3$
$R_3 = 6000\text{ }\Omega - 3300\text{ }\Omega$
$R_3 = 2700\text{ }\Omega = 2.7\text{ k}\Omega$
Step 3: Calculate voltage drops across $R_1$ and $R_2$
Since $I$ is constant at 2.0 mA (0.002 A) throughout the series loop:
$V_1 = I \times R_1 = 0.002\text{ A} \times 1800\text{ }\Omega = 3.6\text{ V}$
$V_2 = I \times R_2 = 0.002\text{ A} \times 1500\text{ }\Omega = 3.0\text{ V}$
Step 4: Calculate power dissipation for $R_3$
Using the shared current and the specific resistance of $R_3$:
$P_3 = I^2 \times R_3$
$P_3 = (0.002\text{ A})^2 \times 2700\text{ }\Omega$
$P_3 = 0.000004\text{ A}^2 \times 2700\text{ }\Omega$
$P_3 = 0.0108\text{ W} = 10.8\text{ mW}$
Sanity Checks and Independent Verification
Before ordering parts, we must verify the math using independent circuit laws. If these checks fail, an algebra error occurred in the steps above.
- KVL Check (Voltage Sum): $V_1 + V_2 + V_3 = 3.6\text{V} + 3.0\text{V} + (0.002\text{A} \times 2700\Omega) = 3.6 + 3.0 + 5.4 = 12.0\text{V}$. This perfectly matches the 12.0V source. Pass.
- Power Sum Check: Total power supplied is $P_{total} = V_{source} \times I_{total} = 12.0\text{V} \times 0.002\text{A} = 24\text{ mW}$. Let's sum the individual dissipations: $P_1 = (0.002)^2 \times 1800 = 7.2\text{ mW}$. $P_2 = (0.002)^2 \times 1500 = 6.0\text{ mW}$. $P_3 = 10.8\text{ mW}$. Sum = $7.2 + 6.0 + 10.8 = 24.0\text{ mW}$. Pass.
- Order of Magnitude Check: Current is in milliamps ($10^{-3}$), resistance is in kilo-ohms ($10^3$). Multiplying them yields volts ($10^0$). The units cancel cleanly without requiring decimal shifts, confirming our 5.4V logic level is in the correct physical ballpark for a 12V system.
Component Selection Decision Tree
Calculating the exact theoretical value is only half the job. Real-world engineering requires mapping theoretical numbers to physical, purchasable components. The E-series standard dictates available resistor values. Below is the decision path to finalize the bill of materials (BOM) for $R_3$.
| Decision Step | Condition / Calculation | Action / Rule | Result |
|---|---|---|---|
| 1. Resistance Value | Calculated: 2700 Ω | Map to nearest E12 standard value | 2.7 kΩ (Exact E12 match) |
| 2. Power Margin | Calculated $P_3$: 10.8 mW | Apply 2x derating safety margin | 21.6 mW minimum rating |
| 3. Wattage Rating | Required: > 21.6 mW | Select smallest standard axial wattage (1/8W = 125mW, 1/4W = 250mW) | 1/8 Watt (125 mW) |
| 4. Tolerance | Feeding logic gate | Requires precision; avoid standard 5% | 1% Tolerance |
Frequently Asked Questions
Why not just use a single 6.0 kΩ resistor instead of three in series?
If you only needed to limit current to 2.0 mA, a single 6.0 kΩ resistor would work perfectly. However, this specific example of a series circuit is designed as a voltage divider. We need the 5.4V tap at the junction between $R_2$ and $R_3$ to feed a microcontroller logic pin. A single resistor cannot provide an intermediate voltage tap.
What happens if the logic gate draws current from the 5.4V tap?
The problem stated the logic circuit is "high-impedance" (drawing microamps or less). If the load draws significant current (e.g., > 0.1 mA), it creates a parallel path with $R_3$. This breaks the pure series model, lowering the equivalent resistance of the bottom leg and causing the tap voltage to sag below 5.4V. For loads drawing more than a few hundred microamps, use a dedicated voltage regulator or an op-amp buffer instead of a passive resistive dropper.
Can I use a 1/4W resistor if I can't find 1/8W in my kit?
Yes. A 1/4W (250 mW) resistor will run significantly cooler than a 1/8W resistor in this circuit because the physical mass is larger and the thermal dissipation margin is massive (250mW rating vs 10.8mW actual). It is electrically identical, just physically bulkier.
